2x - y = 1
\(\Rightarrow\) y = 2x - 1

and x + 2y = 13
\(\Rightarrow \quad y=\frac { 13-x }{ 2 } \)

Plotting the above points and drawing the lines joining them, we get the graph of above equations. Clearly, two lines intersect at point A(3, 5). Hence, x = 3 and .y = 5 is the solution of above equations.
ABC is the triangular shaded region formed by the lines and the Y-axis.

Let the cost of one bat be Rs.x and one ball be Rs.y.
Then, the algebraic representation is given by the following two equations:
3x+6y = 3900 ...(i)
and x+ 3y = 1300 ....(ii)
Now on plotting the graph similar to question 1, we get the following graphical representation.

These two lines intersect at point B (1300, O).Thus, we get the required geometrical representation of given situations.

Fixed charges=Rs.500; cost of food per day=Rs.50

Given linear equations are
x-3y=2 x-3y-2=0 ...(i)
and -2x+6y=5 -2x+6y-5=0 ..(ii)
Here,a₁=1, b₁=-3, c₂=-2
and a₂=-2, b₂=6, c₂=-5
Now, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { 1 }{ -2 } =-\frac { 1 }{ 2 } ,\quad \frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { -3 }{ 6 } =-\frac { 1 }{ 2 } \)
and \(\frac { { c }_{ 1 } }{ { c }_{ 2 } } =\frac { -2 }{ -5 } =\frac { 2 }{ 5 } \)
Here, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
So, the paths represented by the equations are parallel, i.e, not intersect each other.
The values depicted from this action are
(i) awareness about environment.
(ii) participatory role of residents in national activities.

Let the actual speed of the train be x km/h and actual time taken be y h.
\(\because \) Distance=speed x Time
\(\therefore\)Distance=xy km
According to the question,
xy=(x+10)(y-2)
\(\Rightarrow\) xy=xy-2x+10y-20
\(\Rightarrow\) 2x-10y+20=0
\(\Rightarrow\) x-5y+10=0 [dividing both sides by 2]  ...(i)
and xy=(x-10)(y+3)
\(\Rightarrow\) xy=xy+3x-10y-30
\(\therefore\) 3x-10y-30=0  ..(ii)
On multiplyying Eq. (i) by 3 and the n subtracting Eq, (ii) from Eq. (i), we get
3(x-5y+10)-(3x-10y-30)=0
\(\Rightarrow\) -5y=-60
\(\Rightarrow\) y=12
On putting y=12 in Eq. (i), we get
x-5x12+10=0
\(\Rightarrow\)x-60+10=0 \(\Rightarrow\) x=50
Hence, the distanc coverred by the train
=50x12=600 km

Given linear equation is 2x+3y-8=0.   ...(i)
(i) For intersecting lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } \neq \frac { { b }_{ 1 } }{ { b }_{ 2 } } \)
\(\therefore\) Any line intersecting with Eq. (i) may be taken as
3x+2y-9=0 or 3x+2y-7=0 
(ii) For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore\) Any line parallel to Eq. (i) may be taken as
6x+9y+7=0 or 2x+3y-12=0
(iii) For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } =\frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
\(\therefore\) Any line coincident to Eq. (i) may be taken as
4x+6y-16=0
There can be several linear equations in each of (i), (ii) and (iii)

By solving both equations. Find the values of x and y and then put these values in \(y=\lambda x+5\) to get required values of \(\lambda .\)
x=340, y=-165, \(\lambda =-\frac { 1 }{ 2 } \)

Let \(\angle A\) and \(\angle B\) be x⁰ and y⁰ , respectively.
Then, \(\angle C=5\angle B=3(\angle A+\angle B)\)
\(\Rightarrow \quad \angle C=5{ y }^{ 0 }\quad \angle C=3({ x }^{ 0 }+{ y }^{ 0 })\)
\(\Rightarrow \quad \angle C=5{ y }^{ 0 }\) and \(\angle C=3({ x }^{ 0 }+{ y }^{ 0 })\)
\(\Rightarrow \quad 5{ y }^{ 0 }=3({ x }^{ 0 }+{ y }^{ 0 })\quad \)
\(\Rightarrow \quad 2{ y }^{ 0 }-3{ x }^{ 0 }=0\quad \quad \quad ..(i)\)
Also, \(\angle A+\angle B+\angle C={ 180 }^{ 0 }\)
\(\Rightarrow \quad { x }^{ 0 }+{ y }^{ 0 }+5{ y }^{ 0 }={ 180 }^{ 0 }\)
\(\Rightarrow \quad { x }^{ 0 }+6{ y }^{ 0 }={ 180 }^{ 0 }\quad ...(ii)\)
Now, solve Eqs. (i) and (ii),
\(\angle A={ 18 }^{ 0 }\angle B={ 27 }^{ 0 }\angle C={ 135 }^{ 0 }\)

Two straight paths are parallel to each other. Hence, they do not cross each other.

x=-2, y=-1

Let present age of Jacob and his son be x and y, respectively.
According to the question,
Condition I. (x-5)=7(y-5) \(\Rightarrow\)x-7y+30=0 ...(i)
Condition II. (x+5)=3(y+5) \(\Rightarrow\)x-3y+10=0 ...(ii)
Hence, algebraic representation is
x-7y+30=0  \(\Rightarrow\)x-3y+10=0
For graphical representation, draw graphs Eq. (i) and Eq. (ii).

Let the speed of steam be x km/h, then speed of person going upstream=(x+5) km/h and speed of person going downstream=(5-x)h
According to question
\(\frac { 4 }{ 5-x } =3\left( \frac { 4 }{ 5-x } \right) \)
2.4 km/h 
Lete the speed of boat in still water be x km/h and that of stream be y km/h, then according to question
\(\frac { 30 }{ x-y } +\frac { 28 }{ x+y } =7\quad \quad ...(i)\)
and \(\frac { 21 }{ x+y } +\frac { 21 }{ x-y } =5\quad \quad ...(ii)\)  
Solve Eq. (i) and Eq. (ii) to get the speed of boat and speed of stream.
Speed of motorboat in still water=10 km/h
Speed of stream=4 km/h

For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
k=-6

Let x and y be length and breadth of rectangle.
Then, its area=xy
According to the questions,
9x-5)(y+2)=xy-80 \(\Rightarrow\) 2x-5y=-70
(x+10)(y-5)=xy+50  \(\Rightarrow\) -5x+10y=100
Length=40 units, breadth=30 units

Let the two numbers be x and y, then \(\frac { x }{ y } =\frac { 5 }{ 6 } \quad \Rightarrow \quad y=\frac { 6x }{ 5 } \quad ..(i)\)
Also, \(\frac { x-8 }{ y-8 } =\frac { 4 }{ 5 } \quad \Rightarrow \quad 5x-4y=8\quad \quad ..(ii)\)
Draw graphs of Eq. (i) and Eq. (ii) to get the required numbers.
Numbe0 are 49 and 48.